Groups G and G', M is a normal subgroup of G'

We are given:

1. h: G -> G' is a group homomorphism.

2. M is a normal subgroup of G'.

3. N is the set of x in G such that h(x) is in M.

Prove that N is a normal subgroup of G.

It's not that hard for me to prove that N is a subgroup of G--that just comes from the fact that h is a group homomorphism. But I really have no idea how to prove that N is not just a subgroup but a normal subgroup. In fact, these normal subgroup proofs are just plain difficult for me right now. Please help.

Re: Groups G and G', M is a normal subgroup of G'

You need to show that $\displaystyle yxy^{-1}\in N$ for all $\displaystyle y\in G$ and $\displaystyle x\in N$. What happens when you apply the definition of N to $\displaystyle yxy^{-1}$?

Re: Groups G and G', M is a normal subgroup of G'

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**emakarov** You need to show that $\displaystyle yxy^{-1}\in N$ for all $\displaystyle y\in G$ and $\displaystyle x\in N$. What happens when you apply the definition of N to $\displaystyle yxy^{-1}$?

Do I just take h(yxy^-1), which clearly is an element in M, use the homomorphism property to split it up into h(y)h(x)h(y^-1), and I'm pretty much done?

EDIT: Wait...do I need to use the fact that x being in N is equivalent to h(x) being in M?

Re: Groups G and G', M is a normal subgroup of G'

Pretty much. You obviously also need to use the assumption that M is normal, which implies that h(yxy^(-1)) = h(y)h(x)h(y)^(-1) ∈ M since h(x) ∈ M and therefore yxy^(-1) ∈ N.

Re: Groups G and G', M is a normal subgroup of G'

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**emakarov** Pretty much. You obviously also need to use the assumption that M is normal, which implies that h(yxy^(-1)) = h(y)h(x)h(y)^(-1) ∈ M since h(x) ∈ M and therefore yxy^(-1) ∈ N.

So h(y)h(x)h(y)^(-1) ∈ M immediately leads to yxy^-1 ∈ N, right? If so, I'm done.

Re: Groups G and G', M is a normal subgroup of G'

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**phys251** EDIT: Wait...do I need to use the fact that x being in N is equivalent to h(x) being in M?

Yes, you need to use both directions of z ∈ N ⇔ h(z) ∈ M (which holds by definition of N): left-to-right for z = x and right-to-left for z = yxy^(-1).

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**phys251** So h(y)h(x)h(y)^(-1) ∈ M immediately leads to yxy^-1 ∈ N, right?

Yes.

Re: Groups G and G', M is a normal subgroup of G'